RBSE Solutions Class 11 Maths Chapter 13 Measures of Dispersion Exercise 13.1

 

Question 1. Write the formula of quartile deviation.
Answer: The formula provided for the coefficient of quartile deviation is:
Coefficient of quartile deviation \( = \frac { Q_3 - Q_1 }{ Q_3 + Q_1 } \)
Here, \( Q_3 \) represents the upper quartile, and \( Q_1 \) represents the lower quartile.
In simple words: This formula helps us understand how spread out the middle half of our data is, compared to the sum of the upper and lower quartile values. It shows the relative spread of the central data.

🎯 Exam Tip: Always remember that the coefficient of quartile deviation uses both subtraction and addition of the quartiles, while the quartile deviation itself only uses subtraction.

 

Question 2. For any variable series \( Q_1 = 61 \) and \( Q_3 = 121 \). Find the quartile deviation.
Answer:
Given:
Lower quartile \( Q_1 = 61 \)
Upper quartile \( Q_3 = 121 \)
The formula for quartile deviation (Q.D.) is:
Quartile deviation (Q.D.) \( = \frac { Q_3 - Q_1 }{ 2 } \)
Now, substitute the given values into the formula:
\( = \frac { 121 - 61 }{ 2 } \)
\( = \frac { 60 }{ 2 } \)
\( = 30 \)
So, the quartile deviation for this series is 30.
In simple words: To find the quartile deviation, we subtract the smaller quartile from the larger one and then divide the result by two. This tells us the average spread from the median of the middle half of the data.

🎯 Exam Tip: Quartile deviation measures the spread of the central 50% of the data. Make sure to subtract \( Q_1 \) from \( Q_3 \) and divide by 2, not 1.

 

Question 4. Find the range and coefficient of range of the following table

Answer:
From the given table:
The largest value (L) in X is 11.5.
The smallest value (S) in X is 4.5.

First, calculate the Range:
Range \( = L - S \)
\( = 11.5 - 4.5 \)
\( = 7 \)

Next, calculate the Coefficient of Range:
Coefficient of Range \( = \frac { L - S }{ L + S } \)
\( = \frac { 11.5 - 4.5 }{ 11.5 + 4.5 } \)
\( = \frac { 7 }{ 16 } \)
\( = 0.4375 \)
Rounding to two decimal places, the coefficient of range is 0.44.
Thus, the range is 7 and the coefficient of range is 0.44.
In simple words: The range shows the difference between the highest and lowest values in the data. The coefficient of range tells us this difference as a proportion, giving a standardized measure of spread relative to the sum of the highest and lowest values.

🎯 Exam Tip: Remember to identify the absolute maximum and minimum values (L and S) from the variable 'X' for calculating range and its coefficient, not from the frequency 'f'.

 

Question 5. Find the inter quartile range and its coefficient for the following series

Answer:
To find the inter quartile range and its coefficient, we first need to prepare a cumulative frequency (c.f) table:
The total number of terms, \( N = 30 \).

Calculate \( Q_1 \):
\( Q_1 = \text{Value of } (\frac { N + 1 }{ 4 })^{th} \text{ term} \)
\( = \text{Value of } (\frac { 30 + 1 }{ 4 })^{th} \text{ term} \)
\( = \text{Value of } (\frac { 31 }{ 4 })^{th} \text{ term} \)
\( = \text{Value of } (7.75)^{th} \text{ term} \)
The cumulative frequency just greater than 7.75 is 10, which corresponds to X = 1.
So, \( Q_1 = 1 \).

Calculate \( Q_3 \):
\( Q_3 = \text{Value of } 3 (\frac { N + 1 }{ 4 })^{th} \text{ term} \)
\( = \text{Value of } 3 (7.75)^{th} \text{ term} \)
\( = \text{Value of } (23.25)^{th} \text{ term} \)
The cumulative frequency just greater than 23.25 is 25, which corresponds to X = 3.
So, \( Q_3 = 3 \).

Now, calculate the Inter Quartile Range:
Inter Quartile Range \( = Q_3 - Q_1 \)
\( = 3 - 1 \)
\( = 2 \)

Next, calculate the Inter Quartile Range Coefficient:
Inter Quartile Range Coefficient \( = \frac { Q_3 - Q_1 }{ Q_3 + Q_1 } \)
\( = \frac { 3 - 1 }{ 3 + 1 } \)
\( = \frac { 2 }{ 4 } \)
\( = 0.5 \)
Thus, the inter quartile range is 2 and its coefficient is 0.5.
In simple words: We first find the lower and upper quartile values from the data. The difference between these two values gives us the inter quartile range. The coefficient is then calculated by dividing this difference by their sum, which helps compare the spread of data in different series.

🎯 Exam Tip: Remember to calculate cumulative frequency first for discrete series questions. Identify \( Q_1 \) and \( Q_3 \) correctly from the cumulative frequency column before calculating the range and coefficient.

 

Question 6. Find the coefficient of range of the following data

Answer:
From the given data:
The lower limit (S) of the first class interval is 10.
The upper limit (L) of the last class interval is 30.

First, calculate the Range:
Range \( = L - S \)
\( = 30 - 10 \)
\( = 20 \)

Next, calculate the Coefficient of Range:
Coefficient of Range \( = \frac { L - S }{ L + S } \)
\( = \frac { 30 - 10 }{ 30 + 10 } \)
\( = \frac { 20 }{ 40 } \)
\( = \frac { 1 }{ 2 } \)
\( = 0.5 \)
Thus, the coefficient of range for this data is 0.5.
In simple words: For grouped data, the range is the difference between the highest upper class limit and the lowest lower class limit. The coefficient of range then gives this difference as a proportion of their sum, showing how spread out the data is overall.

🎯 Exam Tip: When dealing with class intervals, the lowest value (S) is the lower limit of the first class, and the highest value (L) is the upper limit of the last class. Do not use midpoints or frequencies for Range calculation.

 

Question 8. Find the decile range and percentile range of the following frequency distribution.

Answer:
To find the decile range and percentile range, we first need to prepare a cumulative frequency (c.f) table from the given distribution:
The total frequency \( N = 50 \) (from adding the frequencies up to 40-50, where cumulative frequency reaches 50). The subsequent rows in the source table seem to be from another dataset or are extra. We will use N=50 based on the "9N/10" and "90N/100" calculations.

Calculation of Decile Range:

Calculate \( D_9 \):
First, find the position of the 9th decile:
\( \frac { 9N }{ 10 } = \frac { 9 \times 50 }{ 10 } = \frac { 450 }{ 10 } = 45 \)
The cumulative frequency just greater than 45 is 47. The corresponding class interval is 30-40.
So, for \( D_9 \):
Lower limit of the class \( l = 30 \)
Frequency of the class \( f = 14 \)
Cumulative frequency of the preceding class \( F = 33 \)
Class interval width \( h = 10 \)
The formula for decile is: \( D_k = l + \frac { \frac { kN }{ 10 } - F }{ f } \times h \)
\( D_9 = 30 + \frac { 45 - 33 }{ 14 } \times 10 \)
\( = 30 + \frac { 12 }{ 14 } \times 10 \)
\( = 30 + \frac { 120 }{ 14 } \)
\( = 30 + 8.571 \)
\( D_9 = 38.571 \)

Calculate \( D_1 \):
First, find the position of the 1st decile:
\( \frac { 1N }{ 10 } = \frac { 1 \times 50 }{ 10 } = 5 \)
The cumulative frequency exactly equal to 5 is 5. The corresponding class interval is 0-10.
So, for \( D_1 \):
Lower limit of the class \( l = 0 \)
Frequency of the class \( f = 5 \)
Cumulative frequency of the preceding class \( F = 0 \) (since no class before 0-10)
Class interval width \( h = 10 \)
\( D_1 = 0 + \frac { 5 - 0 }{ 5 } \times 10 \)
\( = 0 + \frac { 5 }{ 5 } \times 10 \)
\( = 0 + 1 \times 10 \)
\( D_1 = 10 \)

Now, calculate the Decile Range:
Decile Range \( = D_9 - D_1 \)
\( = 38.571 - 10 \)
\( = 28.571 \)

Calculation of Percentile Range:

Calculate \( P_{90} \):
First, find the position of the 90th percentile:
\( \frac { 90N }{ 100 } = \frac { 90 \times 50 }{ 100 } = \frac { 4500 }{ 100 } = 45 \)
The cumulative frequency just greater than 45 is 47. The corresponding class interval is 30-40.
So, for \( P_{90} \):
Lower limit of the class \( l = 30 \)
Frequency of the class \( f = 14 \)
Cumulative frequency of the preceding class \( F = 33 \)
Class interval width \( h = 10 \)
The formula for percentile is: \( P_k = l + \frac { \frac { kN }{ 100 } - F }{ f } \times h \)
\( P_{90} = 30 + \frac { 45 - 33 }{ 14 } \times 10 \)
\( = 30 + \frac { 12 }{ 14 } \times 10 \)
\( = 30 + \frac { 120 }{ 14 } \)
\( = 30 + 8.571 \)
\( P_{90} = 38.571 \)

Calculate \( P_{10} \):
First, find the position of the 10th percentile:
\( \frac { 10N }{ 100 } = \frac { 10 \times 50 }{ 100 } = 5 \)
The cumulative frequency exactly equal to 5 is 5. The corresponding class interval is 0-10.
So, for \( P_{10} \):
Lower limit of the class \( l = 0 \)
Frequency of the class \( f = 5 \)
Cumulative frequency of the preceding class \( F = 0 \)
Class interval width \( h = 10 \)
\( P_{10} = 0 + \frac { 5 - 0 }{ 5 } \times 10 \)
\( = 0 + \frac { 5 }{ 5 } \times 10 \)
\( = 0 + 1 \times 10 \)
\( P_{10} = 10 \)

Now, calculate the Percentile Range:
Percentile Range \( = P_{90} - P_{10} \)
\( = 38.571 - 10 \)
\( = 28.571 \)
Thus, the decile range is 28.571 and the percentile range is 28.571.
In simple words: The decile range shows the spread between the 10th and 90th deciles, which divide the data into ten equal parts. The percentile range is similar but uses the 10th and 90th percentiles, which divide data into one hundred equal parts. Both help understand the spread of the middle 80% of the data.

🎯 Exam Tip: When calculating deciles and percentiles for grouped data, always construct a cumulative frequency table first. Ensure you correctly identify the relevant class interval, lower limit (l), frequency (f), and preceding cumulative frequency (F) for each calculation.